Dimension of R over Q without cardinality argument. [duplicate]

I am looking for the easiest (elementary) proof that $\mathbb R$ is infinite dimensional as a $\mathbb Q$-vector space, without using cardinality. It should be understandable at highschool level.

So I guess the question could be reformulated as: what is the easiest infinite family of reals that can be showed to be independent ? So far square roots of primes seems a good candidate, but the proof is still a little intricate, is it the easiest possible ?

The goal of this is to show students that we can prove the result by different ways, and see that understanding cardinals is useful. But I still want the students to be able to understand the other proof.

Solution 1:

Consider the set:

$$\{\log p\}_p$$

of logs of prime numbers. They are linearly independent over $\Bbb Q$ by the Fundamental Theorem of Arithmetic--i.e. unique factorization of integers greater than $1$ into primes. As there are infinitely many primes, the set $\Bbb R$ contains an infinite dimensional subspace, and is so itself infinite dimensional. (This is the standard proof number theorists love).

Solution 2:

Try $\{1, \pi, \pi^2, \pi^3,\ldots\}$ (or use any transcendental number instead of $\pi$). Since if $$\sum_{i=0}^na_i\pi^i=0, \text{ }a_i\in\mathbb Q$$then $\pi$ must be a root of $$a_0 + a_1x + \ldots + a_nx^n$$which would contradict the transcendence of $\pi$.